Simplify and expand the following expression: $ \dfrac{3}{5z + 25}- \dfrac{3}{z - 2}- \dfrac{4}{z^2 + 3z - 10} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{3}{5z + 25} = \dfrac{3}{5(z + 5)}$ We can factor the quadratic in the third term: $ \dfrac{4}{z^2 + 3z - 10} = \dfrac{4}{(z + 5)(z - 2)}$ Now we have: $ \dfrac{3}{5(z + 5)}- \dfrac{3}{z - 2}- \dfrac{4}{(z + 5)(z - 2)} $ The least common multiple of the denominators is: $ 5(z + 5)(z - 2)$ In order to get the first term over $5(z + 5)(z - 2)$ , multiply by $\dfrac{z - 2}{z - 2}$ $ \dfrac{3}{5(z + 5)} \times \dfrac{z - 2}{z - 2} = \dfrac{3(z - 2)}{5(z + 5)(z - 2)} $ In order to get the second term over $5(z + 5)(z - 2)$ , multiply by $\dfrac{5(z + 5)}{5(z + 5)}$ $ \dfrac{3}{z - 2} \times \dfrac{5(z + 5)}{5(z + 5)} = \dfrac{15(z + 5)}{5(z + 5)(z - 2)} $ In order to get the third term over $5(z + 5)(z - 2)$ , multiply by $\dfrac{5}{5}$ $ \dfrac{4}{(z + 5)(z - 2)} \times \dfrac{5}{5} = \dfrac{20}{5(z + 5)(z - 2)} $ Now we have: $ \dfrac{3(z - 2)}{5(z + 5)(z - 2)} - \dfrac{15(z + 5)}{5(z + 5)(z - 2)} - \dfrac{20}{5(z + 5)(z - 2)} $ $ = \dfrac{ 3(z - 2) - 15(z + 5) - 20} {5(z + 5)(z - 2)} $ Expand: $ = \dfrac{3z - 6 - 15z - 75 - 20}{5z^2 + 15z - 50} $ $ = \dfrac{-12z - 101}{5z^2 + 15z - 50}$